Metamath Proof Explorer


Theorem s4prop

Description: A length 4 word is a union of two unordered pairs of ordered pairs. (Contributed by Alexander van der Vekens, 14-Aug-2017)

Ref Expression
Assertion s4prop ASBSCSDS⟨“ABCD”⟩=0A1B2C3D

Proof

Step Hyp Ref Expression
1 df-s4 ⟨“ABCD”⟩=⟨“ABC”⟩++⟨“D”⟩
2 simpl ASBSAS
3 2 adantr ASBSCSDSAS
4 simpr ASBSBS
5 4 adantr ASBSCSDSBS
6 simpl CSDSCS
7 6 adantl ASBSCSDSCS
8 3 5 7 s3cld ASBSCSDS⟨“ABC”⟩WordS
9 simpr CSDSDS
10 9 adantl ASBSCSDSDS
11 cats1un ⟨“ABC”⟩WordSDS⟨“ABC”⟩++⟨“D”⟩=⟨“ABC”⟩⟨“ABC”⟩D
12 8 10 11 syl2anc ASBSCSDS⟨“ABC”⟩++⟨“D”⟩=⟨“ABC”⟩⟨“ABC”⟩D
13 df-s3 ⟨“ABC”⟩=⟨“AB”⟩++⟨“C”⟩
14 s2cl ASBS⟨“AB”⟩WordS
15 14 adantr ASBSCSDS⟨“AB”⟩WordS
16 cats1un ⟨“AB”⟩WordSCS⟨“AB”⟩++⟨“C”⟩=⟨“AB”⟩⟨“AB”⟩C
17 15 7 16 syl2anc ASBSCSDS⟨“AB”⟩++⟨“C”⟩=⟨“AB”⟩⟨“AB”⟩C
18 13 17 eqtrid ASBSCSDS⟨“ABC”⟩=⟨“AB”⟩⟨“AB”⟩C
19 s2prop ASBS⟨“AB”⟩=0A1B
20 19 adantr ASBSCSDS⟨“AB”⟩=0A1B
21 20 uneq1d ASBSCSDS⟨“AB”⟩⟨“AB”⟩C=0A1B⟨“AB”⟩C
22 18 21 eqtrd ASBSCSDS⟨“ABC”⟩=0A1B⟨“AB”⟩C
23 22 uneq1d ASBSCSDS⟨“ABC”⟩⟨“ABC”⟩D=0A1B⟨“AB”⟩C⟨“ABC”⟩D
24 12 23 eqtrd ASBSCSDS⟨“ABC”⟩++⟨“D”⟩=0A1B⟨“AB”⟩C⟨“ABC”⟩D
25 unass 0A1B⟨“AB”⟩C⟨“ABC”⟩D=0A1B⟨“AB”⟩C⟨“ABC”⟩D
26 25 a1i ASBSCSDS0A1B⟨“AB”⟩C⟨“ABC”⟩D=0A1B⟨“AB”⟩C⟨“ABC”⟩D
27 df-pr ⟨“AB”⟩C⟨“ABC”⟩D=⟨“AB”⟩C⟨“ABC”⟩D
28 s2len ⟨“AB”⟩=2
29 28 a1i ASBSCSDS⟨“AB”⟩=2
30 29 opeq1d ASBSCSDS⟨“AB”⟩C=2C
31 s3len ⟨“ABC”⟩=3
32 31 a1i ASBSCSDS⟨“ABC”⟩=3
33 32 opeq1d ASBSCSDS⟨“ABC”⟩D=3D
34 30 33 preq12d ASBSCSDS⟨“AB”⟩C⟨“ABC”⟩D=2C3D
35 27 34 eqtr3id ASBSCSDS⟨“AB”⟩C⟨“ABC”⟩D=2C3D
36 35 uneq2d ASBSCSDS0A1B⟨“AB”⟩C⟨“ABC”⟩D=0A1B2C3D
37 24 26 36 3eqtrd ASBSCSDS⟨“ABC”⟩++⟨“D”⟩=0A1B2C3D
38 1 37 eqtrid ASBSCSDS⟨“ABCD”⟩=0A1B2C3D